December 17, 2018 December 17, 2018 06:00 PM PST 07:00 PM PST Join our live webinar and learn how to approach Data Sufficiency and Critical Reasoning problems, how to identify the best way to solve each question and what most people do wrong. December 17, 2018 December 17, 2018 10:00 PM PST 11:00 PM PST From Dec 5th onward, American programs will start releasing R1 decisions. Chat Rooms: We have also assigned chat rooms for every school so that applicants can stay in touch and exchange information/update during decision period.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 26 Apr 2010
Posts: 108
Concentration: Strategy, Entrepreneurship

If p+q = j and pq = k, then 2pq =
[#permalink]
Show Tags
26 Nov 2010, 12:11
Question Stats:
76% (01:50) correct 24% (02:38) wrong based on 329 sessions
HideShow timer Statistics
If p+q = j and pq = k, then 2pq = A. (j^2  k^2)/2 B. (k^2  j^2)/2 C. jk/2 D. (j^2 + k^2)/2 E. (j  k)/2 Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically... Can someone post the algebraic solution? Thanks in advance.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
I appreciate the kudos if you find this post helpful! +1



Math Expert
Joined: 02 Sep 2009
Posts: 51259

Re: Solving for 2pq
[#permalink]
Show Tags
26 Nov 2010, 12:21



Manager
Joined: 26 Apr 2010
Posts: 108
Concentration: Strategy, Entrepreneurship

Re: Solving for 2pq
[#permalink]
Show Tags
26 Nov 2010, 12:43
Thanks very much Bunuel. +1
_________________
I appreciate the kudos if you find this post helpful! +1



Manager
Joined: 13 Jul 2010
Posts: 136

Re: Solving for 2pq
[#permalink]
Show Tags
26 Nov 2010, 13:03
Bunuel wrote: martie11 wrote: Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...
If p+q = j and pq = k, then 2pq =
a) (j^2  k^2)/2 b) (k^2  j^2)/2 c) jk/2 d) (j^2 + k^2)/2 e) (j  k)/2
Can someone post the algebraic solution? Thanks in advance. Given: \(p+q = j\) and \(pq = k\) > square both expressions: \(p^2+2pq+q^2=j^2\) and \(p^22pq+q^2=k^2\) > now, subtract (2) form (1): \(4pq=j^2k^2\) > \(2pq=\frac{j^2k^2}{2}\). Answer: A. Bunuel did you know to square all of the terms once you looked at the answers  since they were all in J^2 or K^2 form? I went the route of understanding this was a play on the a^2 +2ab+b^2 form but wouldn't think to subtract the two unless I looked at the answer choices. Just wanted to understand your approach.



Manager
Joined: 26 Apr 2010
Posts: 108
Concentration: Strategy, Entrepreneurship

Re: Solving for 2pq
[#permalink]
Show Tags
26 Nov 2010, 13:23
Good question...my thought exactly...I didn't 'see' that I should square both sides...
_________________
I appreciate the kudos if you find this post helpful! +1



Intern
Joined: 03 Jun 2009
Posts: 45

Re: Solving for 2pq
[#permalink]
Show Tags
27 Nov 2010, 09:49
gettinit wrote: Bunuel wrote: martie11 wrote: Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...
If p+q = j and pq = k, then 2pq =
a) (j^2  k^2)/2 b) (k^2  j^2)/2 c) jk/2 d) (j^2 + k^2)/2 e) (j  k)/2
Can someone post the algebraic solution? Thanks in advance. Given: \(p+q = j\) and \(pq = k\) > square both expressions: \(p^2+2pq+q^2=j^2\) and \(p^22pq+q^2=k^2\) > now, subtract (2) form (1): \(4pq=j^2k^2\) > \(2pq=\frac{j^2k^2}{2}\). Answer: A. Bunuel did you know to square all of the terms once you looked at the answers  since they were all in J^2 or K^2 form? I went the route of understanding this was a play on the a^2 +2ab+b^2 form but wouldn't think to subtract the two unless I looked at the answer choices. Just wanted to understand your approach. These kind of questions wud b easier if we make a note that.... (a+b)^2 = a^2 + b^2 + 2ab (ab)^2 = a^2 + b^2 2ab => a^2 + b^2 = [(a+b)^2 + (ab)^2]/2...Sum of the squares *0.5 also => 4ab = [(a+b)^2  (ab)^2]/2 ...difference of the squares *0.5



Manager
Joined: 01 Nov 2010
Posts: 126
Location: Zürich, Switzerland

Re: Solving for 2pq
[#permalink]
Show Tags
Updated on: 29 Nov 2010, 15:22
Look right at 2pq which hints at use of the formula (x+y)2 and (xy)2
subtracting (xy)2 from (x+y)2 above gives 4pq which when divided by 2 gives the result.
Answer: A
Originally posted by Sarang on 27 Nov 2010, 17:46.
Last edited by Sarang on 29 Nov 2010, 15:22, edited 1 time in total.



Manager
Joined: 30 Aug 2010
Posts: 88
Location: Bangalore, India

Re: Solving for 2pq
[#permalink]
Show Tags
29 Nov 2010, 06:27
martie11 wrote: Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...
If p+q = j and pq = k, then 2pq =
a) (j^2  k^2)/2 b) (k^2  j^2)/2 c) jk/2 d) (j^2 + k^2)/2 e) (j  k)/2
Can someone post the algebraic solution? Thanks in advance. MAKE IT SIMPLE. ADD 1st equation to second ==> 2p=j+k SUBTRACT 2nd equation from 1st ==> 2q=jk multiply the above 2 equations 4pq=j^2K^2 ==> 2pq = j^2K^2 / 2 ANSWER "A" Regards, Murali. Kudos?



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1825
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: If p+q = j and pq = k, then 2pq =
[#permalink]
Show Tags
17 Jul 2014, 01:53
p+q = j ........... (1) pq = k ............ (2) (1) + (2) \(p = \frac{j+k}{2}\) (1)  (2) \(q = \frac{jk}{2}\) \(2pq = 2 * \frac{j+k}{2} *\frac{jk}{2}\) \(= \frac{j^2  k^2}{2}\) Answer = A
_________________
Kindly press "+1 Kudos" to appreciate



Intern
Joined: 14 Aug 2016
Posts: 1
Location: India
Concentration: International Business, Finance
WE: Consulting (Consulting)

Re: If p+q = j and pq = k, then 2pq =
[#permalink]
Show Tags
26 Oct 2016, 23:06
martie11 wrote: If p+q = j and pq = k, then 2pq =
A. (j^2  k^2)/2 B. (k^2  j^2)/2 C. jk/2 D. (j^2 + k^2)/2 E. (j  k)/2
Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...
Can someone post the algebraic solution? Thanks in advance. here you go, (p+q)^2 = p^2 + q^2 +2pq = j^2 ................1 (pq)^2 = p^2 + q^2 2pq = k^2 ..................2 subtract 2 from 1 4pq=j^2  k^2 ........................3 divide eqaution 3 by 2 on both sides 2pq = (j^2  k^2)/2 which is option A.



NonHuman User
Joined: 09 Sep 2013
Posts: 9197

Re: If p+q = j and pq = k, then 2pq =
[#permalink]
Show Tags
29 Jun 2018, 21:45
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: If p+q = j and pq = k, then 2pq = &nbs
[#permalink]
29 Jun 2018, 21:45






